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$f:(0,1) \rightarrow \mathbb{R}$, and for every Cauchy sequence $x_n \in (0,1)$, $(f \circ x)_n$ is Cauchy: does that mean that $f$ is uniformly continuous ?

I believe the answer is yes, continuity has already been asked and answered. But what about uniform continuity?

In an exercise, $x^2$ was given as a counterexample. However, I believe this to be incorrect on $(0,1)$.

I belive the implication in the Title to be correct. Here is my attempt at a proof:

Consider $[0,1]$, the closure of $(0,1)$. It is complete and thus, any Cauchy sequence in $[0,1]$ also converges in $[0,1]$. Consider $x^{(l)}_n \rightarrow l$, $l \in \{0,1\}$. Since $f(x_n)$ is Cauchy, $f(x^{(l)}_n) \rightarrow L^{(l)} < \infty$. We can therefore naturally extend $$\tilde{f}:[0,1] \rightarrow \mathbb{R}, x \mapsto \begin{cases} L^{(l)} & x = l \\ f(x) & x \in (0,1) \end{cases}$$ Heine's Theorem states that a continuous function on a compact is uniformly continuous and therefore $\tilde{f}$ is uniformly continuous and so is $f$ (you can just take off $\{0,1\}$ in the $\varepsilon,\delta$ definition of $\tilde{f}$).

Is that correct?

P.S. The comment by Mark points out a step that I overlooked.

Pastudent
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  • True. $x\mapsto x^2$ is uniformly continuous on $(0,1)$, yes – FShrike Feb 10 '23 at 18:59
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    @FShrike I guess we can keep deleting comments :D – Pastudent Feb 10 '23 at 18:59
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    You need to be careful. You extended $f$ using a limit of just one sequence. How can you be sure that if you take a different sequence $y_n$ which tends to $l$ then the limit of $f(y_n)$ will still be the same $L^{l}$? Because otherwise, you don't get a continuous function in $[0,1]$. – Mark Feb 10 '23 at 19:00
  • @Mark You are right. Indeed, I am implicitely using the fact that $f$ is continuous (which was proved in a linked post). – Pastudent Feb 10 '23 at 19:04
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    @Pastudent It is continuous in $(0,1)$, you wish to extend it continuously to $[0,1]$ as well. The statement is true anyway. Thing is, if $y_n\to l$ then you can look at the sequence $x_1,y_1,x_2,y_2,x_3,y_3,...$. It also converges to $l$, and so $f(x_1), f(y_1), f(x_2), f(y_2),...$ is Cauchy, and hence converges in $\mathbb{R}$. Since it contains $f(x_n)$ and $f(y_n)$ as subsequences we conclude that $f(x_n)$ and $f(y_n)$ must have the same limit. – Mark Feb 10 '23 at 19:08
  • @Mark I guess you could write your whole point as an answer. – Pastudent Feb 10 '23 at 19:09
  • @Pastudent Ok, I added an answer. – Mark Feb 10 '23 at 19:15

1 Answers1

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The statement is true, but the proof needs to be fixed a bit. Thing is, there exists a real number $L_0\in\mathbb{R}$ such that if $x_n$ is any sequence which tends to $0$ then $f(x_n)\to L_0$. You already proved that if $x_n\to 0$ then the limit of $f(x_n)$ exists. Now we just need to show this limit does not depend on the chosen sequence $x_n$. So assume $y_n\to 0$ as well. Then consider the sequence:

$x_1,y_1,x_2,y_2,...$

It tends to $0$ as well, and so $f(x_1), f(y_1), f(x_2), f(y_2),...$ is Cauchy, and hence converges in $\mathbb{R}$. Since it contains $f(x_n)$ and $f(y_n)$ as subsequences, we deduce that $\lim\limits_{n\to\infty} f(x_n)=\lim\limits_{n\to\infty} f(y_n)$.

Similarly, there is some $L_1$ such that if $x_n\to 1$ then $f(x_n)\to L_1$. And now we can indeed extend $f$ to a continuous function on $[0,1]$ by letting $f(0)=L_0$ and $f(1)=L_1$. Since $[0,1]$ is compact, it is uniformly continuous.

Mark
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